Two small steel balls A and B have mass 0.6 kg and 0.2 kg respectively. They are moving towards each other in opposite directions on a smooth horizontal table when they collide directly. Immediately before the collision, the speed of A is 8 m s⁻¹ and the speed of B is 2 m s⁻¹. Immediately after the collision, the direction of motion of A is unchanged and the speed of B is twice the speed of A. Find (a) the speed of A immediately after the collision, (b) the magnitude of the impulse exerted on B in the collision.

A-Level Edexcel Maths Mechanics Forces: Two small steel balls A and B ha

Two small steel balls A and B have mass 0.6 kg and 0.2 kg respectively - Edexcel - A-Level Maths Mechanics - Question 2 - 2005 - Paper 1

Question 2

Two small steel balls A and B have mass 0.6 kg and 0.2 kg respectively. They are moving towards each other in opposite directions on a smooth horizontal table when t... show full transcript

Worked Solution & Example Answer:Two small steel balls A and B have mass 0.6 kg and 0.2 kg respectively - Edexcel - A-Level Maths Mechanics - Question 2 - 2005 - Paper 1

Step 1

the speed of A immediately after the collision

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Answer

To solve for the speed of A immediately after the collision, we will use the principle of conservation of linear momentum:

$m_A v_A + m_B v_B = m_A v'_{A} + m_B v'_{B}$

Substituting the known values:

Mass of A, $m_A = 0.6 ext{ kg}$
Mass of B, $m_B = 0.2 ext{ kg}$
Speed of A before collision, $v_A = 8 ext{ m s}^{-1}$
Speed of B before collision, $v_B = -2 ext{ m s}^{-1}$ (negative because it moves in the opposite direction)
Let the speed of A after the collision be $v$ and the speed of B after the collision be $2v$ .

Plugging these into the momentum equation gives:

$0.6 imes 8 + 0.2 imes (-2) = 0.6v + 0.2(2v)$

Solving this results in:

$4.8 - 0.4 = 0.6v + 0.4v$

Which simplifies to:

$4.4 = v$

Thus, the speed of A after the collision is $4.4 ext{ m s}^{-1}$ .

Step 2

the magnitude of the impulse exerted on B in the collision

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Answer

To find the impulse exerted on B, we utilize the impulse-momentum theorem, which states:

$J = ext{Change in momentum} = m_B (v'_{B} - v_B)$

Substituting the known values:

Mass of B, $m_B = 0.2 ext{ kg}$
Speed of B before the collision, $v_B = -2 ext{ m s}^{-1}$
Speed of B after the collision, $v'_{B} = 2v = 2(4.4) = 8.8 ext{ m s}^{-1}$ .

Thus, the impulse is calculated as follows:

$J = 0.2(8.8 - (-2)) = 0.2(8.8 + 2) = 0.2(10.8) = 2.16 ext{ Ns}$

Therefore, the magnitude of the impulse exerted on B during the collision is $2.16 ext{ Ns}$ .

Two small steel balls A and B have mass 0.6 kg and 0.2 kg respectively - Edexcel - A-Level Maths Mechanics - Question 2 - 2005 - Paper 1

Worked Solution & Example Answer:Two small steel balls A and B have mass 0.6 kg and 0.2 kg respectively - Edexcel - A-Level Maths Mechanics - Question 2 - 2005 - Paper 1

the speed of A immediately after the collision

the magnitude of the impulse exerted on B in the collision

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